On Thu, Aug 28, 2003 at 11:21:08PM -0700, Sarad AV wrote: > hi, > > Let ~ represents a relation. > > If a~b and b~a,then > > a~a (by transitivity) > is an incorrect argument.
> By definition of transitivity, if a~b and b~c implies > that a~c. right. > I was asking on the same lines if (a*d)*d=a*(d*d)=d. What does that have to do with transitivity? You didn't mention transitivity when you posed the question. Ridiculous. '*' is an operator, not a relation. Relations can't be parenthesized unless you're going to make truth or falsehood a symbol to be operated upon. Tim is right. Cypherpunks isn't a place to look for help with your algebra homework. I like doing interesting math problems, but you're not even properly asking the questions you want answered. That makes it a LOT less fun. > By definition associativity is defined on a,b,c > element of set S and not two elements of the set. This is getting stupid. The '*' operator was defined associative. The property of associativity applies to ASSOCIATIONS between symbols (i.e binary operators). > x*y (ie, left*top) can be followed. I'm totally done with this.
